Question

4 tan x-
1/2 sec x + 2 sin x --
3 / sin x + 6 sec x /cosec x + 5e^x

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Answer 1

$\frac{d}{dx}[ 4 tan x-\frac{sec x }{2} + 2 sin x -\frac{3}{sin x} + 6\frac{sec x}{cosec x} + 5e^x]$

⇒ $4.[\frac{d}{dx} [tan x]]+ 2.[\frac{d}{dx} [sin x]]+3.[\frac{d}{dx} [\frac{1}{sin x} ]]+6[\frac{d}{dx} [\frac{sec x}{cosec x} ]] - \frac{1}{2} [\frac{d}{dx} sec x]+5[\frac{d}{dx} [e^x]]$

⇒ $4 sec^{2} x +2cos x - 3.\frac{\frac{d}{dx} sinx}{sin^{2} x} + 6.\frac{\frac{d}{dx} [sec x. cosec x- secx\frac{d}{dx}cosec x] }{cosec^{2} x} - \frac{secxtanx}{2} + 5e^x$

⇒ $4 sec^{2} x +2cos x - 3.\frac{cosx}{sin^{2} x} + 6.\frac{ [sec x. tanx.cosec x- secx.(-cotx).cosec x] }{cosec^{2} x} - \frac{secxtanx}{2} + 5e^x$

⇒ $4 sec^{2} x +2cos x - 3.\frac{cosx}{sin^{2} x} + 6.\frac{ [sec x. tanx.cosec x+ secx.cotx.cosec x] }{cosec^{2} x} - \frac{secxtanx}{2} + 5e^x$

⇒  $6.\frac{ sec x. tanx}{cosecx}- \frac{secxtanx}{2}- 3.\frac{cosx}{sin^{2} x}+4 sec^{2} x+6.\frac{ secx.cotx}{cosec x} +2cos x + 5e^x$

⇒ $10 tanx + 2 sin x -\frac{secx}{2} +3cosec x+5e^x$